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Chapter 13

50.00 g of a solid at 25.00 o C is placed in 250.00 g of its liquid in a 500.00 g can at 77.00 o C. The temp. equilibrates at 27.00 o C. Calculate: H f of the solid C liquid = 2.00 J/gK C can = 1.00 J/gK MP solid = 25.00 o C. Chapter 13. States of Matter. Fluid.

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Chapter 13

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  1. 50.00 g of a solid at 25.00oC is placed in 250.00 g of its liquid in a 500.00 g can at 77.00oC. The temp. equilibrates at 27.00oC. Calculate: Hf of the solidCliquid = 2.00 J/gK Ccan = 1.00 J/gKMPsolid = 25.00oC

  2. Chapter 13 States of Matter

  3. Fluid • Any substance that flow and has no definite shape

  4. Fluids • Both liquids & gases are fluids

  5. Pressure • Applying a force to a surface results in pressure

  6. Pressure (P) Force per unit area P = F/A

  7. Common Units of P Atm Torcelli mm Hg inches Hg Bar millibar Pascal kilopascal

  8. Std Unit for Pressure Kilopascal (kPa) Pa = N/m2

  9. Standard Pressure 101.3 kPa 760 mmHg ~ 1 x 105 Pa 1.0 Atm 30 in Hg 1013 mbar

  10. If std. air pressure = 101.3 kPa, calculate the force on a desk top that is 50.0 cm x 100.0 cm.

  11. Each car tire makes contact with the ground on an area that is 5.0 x 6.0 inches. The tire pressure is 40 psi. Calculate the weight of the car.

  12. Fluids at Rest

  13. Pascal’s Principle • Pressure applied at any point in a confined fluid is transmitted undiminished throughout the fluid

  14. Pressure Formula F A P =

  15. In Constant Pressure P1 = P2

  16. Thus F1 F2 A1 A2 =

  17. Thus F1A2 A1 F2 =

  18. Thus A2 A1 F2=F1()

  19. Using a car jack a person applies 50.0 lbs force on a 2.0 mm wide cylinder driving the fluid into a 2.0 cm wide piston. Calculate the force applied to the piston.

  20. Density (r) • Mass/unit volume m V r =

  21. In a fork lift, fluid is pumped through a 0.20 mm tube with a force of 15 N. The small tube opens up into a 20.0 cm cylinder driving a piston with ___ N

  22. Gases

  23. Boyle’s Law • At constant temperature, pressure & volume are inversely proportioned

  24. Boyle’s Law P1V1 = P2V2

  25. Charles’ Law • At constant pressure, volume & temperature are directly proportioned

  26. Charles’ Law V1 V2 T1 T2 =

  27. Guy L’ Law • At constant volume, pressure & temperature are directly proportioned

  28. GL’ Law P1 P2 T1 T2 =

  29. C G Law P1V1 P2V2 T1 T2 =

  30. Ideal Gas Law PV = nRT

  31. Ideal Gas Constant 8.31 LkPa moleK R =

  32. Ideal Gas Constant 8.31 J moleK R =

  33. Ideal Gas Constant 0.0821 LAtm moleK R=

  34. Calculate the new volume if the conditions of 20.0 L gas at 27oC under 150 kPa is changed to 177oC under 180 kPa:

  35. Calculate the change in volume if the pressure on a gas is halved while its temperature is tripled:

  36. Calculate the volume 0.831 moles of gas at 27oC under 150 kPa: R = 8.31 LkPa/moleK

  37. Calculate the number of moles of gas occupying 831 mL at 77oC under 175 kPa: R = 8.31 LkPa/moleK

  38. Pressure in Water • The pressure applied by the weight of water pressing down on an area

  39. Pressure Formula W A P =

  40. W = mg, thus: mg A P =

  41. m = rV, thus: rVg A P =

  42. V = Ah, thus: rAhg A P =

  43. A’s cancel, thus: P = rhg

  44. The Pressure of a Body of Water P = rhg

  45. The Pressure of a Body of Water Because r & g are constant, P depends on only the height

  46. Buoyant Force • The upward force applied by water due to the pressure difference between the top & bottom of a submerged object

  47. Buoyant Force • Ftop = PtopA = rhgA • Fbtm = PbtmA = r(h + l)gA

  48. Buoyant Force • Fbuoy = Fbtm - Ftop • Fbuoy = r(h + l)gA - rhgA • Fbuoy = r lgA = rVg

  49. Buoyant Force • Fbuoy = rVg • Fbuoy = Wwater

  50. Archimedes’ Principle • A object immersed in fluid has an upward force equal to the weight of fluid displaced

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